K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 12 2017

làm hộ mình cái để mai nộp thầy,ai nhanh và đúng thì mình k cho nha

13 tháng 4 2019

\(a)-3\frac{1}{2}+\frac{1}{3}.\left(x-1\right)=-1\frac{1}{3}:2\frac{1}{3}\)

\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{3}:\frac{7}{3}\)

\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}\)

\(\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}-\frac{-7}{2}\)

\(\frac{1}{3}.\left(x-1\right)=\frac{41}{14}\)

\(\Rightarrow x-1=\frac{41}{14}:\frac{1}{3}\)

\(\Rightarrow x-1=\frac{123}{14}\)

\(\Rightarrow x=\frac{123}{14}+1\)

\(\Rightarrow x=\frac{137}{14}\)

Bài 1: Thu gọn a) \(\frac{1}{5}x^4y^3-3x^4y^3\) b) \(5x^2y^5-\frac{1}{4}x^2y^5\) c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\) d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\) e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\) f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\) g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\) h)...
Đọc tiếp

Bài 1: Thu gọn

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)

d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)

e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)

f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)

g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)

h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)

i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)

k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)

n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)

m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)

p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)

1
26 tháng 7 2019

Bài 1:

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

= \(\left(\frac{1}{5}-3\right)x^4y^3\)

= \(-\frac{14}{5}x^4y^3.\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

= \(\left(5-\frac{1}{4}\right)x^2y^5\)

= \(\frac{19}{4}x^2y^5.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!

29 tháng 7 2019

cảm ơn nha

chúc bạn học tốt

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

26 tháng 10 2016

a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

        \(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

        \(\frac{2}{5}-x=-3\)

                   \(x=\frac{2}{5}-\left(-3\right)\)

                   \(x=\frac{2}{5}+3\)

                   \(x=\frac{3}{5}-\frac{15}{5}\)

                   \(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\) 

    

  

26 tháng 10 2016

b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

        \(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

        \(-3+\frac{3}{x}=\frac{-25}{12}\)

                     \(\frac{3}{x}=\frac{-25}{12}+3\)

                      \(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

                      \(\frac{3}{x}=\frac{5}{6}\)

                      \(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ

16 tháng 8 2019

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

16 tháng 8 2019

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

8 tháng 10 2019

a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)

\(\frac{1}{2}-x=\frac{57}{28}\)

\(x=-\frac{43}{28}\)

b, \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow2x-1=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

8 tháng 10 2019

b, \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow\left(2x-1\right)^2=5^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy ...

8 tháng 10 2019

a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)

\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)

\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)

\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)

\(\Rightarrow x=-\frac{43}{28}\)

Vậy \(x=-\frac{43}{28}.\)

b) \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=20+5\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow2x-1=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{3;-2\right\}.\)

d) \(\frac{x-6}{4}=\frac{4}{x-6}\)

\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)

\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)

\(\Rightarrow\left(x-6\right)^2=16\)

\(\Rightarrow x-6=\pm4\)

\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{10;2\right\}.\)

Chúc bạn học tốt!